Question

The least positive integer $n$ such that ${\left[\begin{array}{cc}\cos \frac{\pi }{4}& \sin \frac{\pi }{4}\\ -\sin \frac{\pi }{4}& \cos \frac{\pi }{4}\end{array}\right]}^n$ is an identity matrix of order $2$ is

• A. $4$
• B. $8$
• C. $12$
• D. $16$

Answer 1

The correct option is B $8$

$\left[\begin{array}{cc}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right]$
Let $A=\left[\begin{array}{cc}\lambda & \lambda \\ -\lambda & \lambda \end{array}\right]$ where $\lambda =\frac{1}{\sqrt{2}}$
$A^2=\left[\begin{array}{cc}\lambda & \lambda \\ -\lambda & \lambda \end{array}\right]\left[\begin{array}{cc}\lambda & \lambda \\ -\lambda & \lambda \end{array}\right]=\left[\begin{array}{cc}0& 2{\lambda }^2\\ -2{\lambda }^2& 0\end{array}\right]$
$A^4=\left[\begin{array}{cc}0& 2{\lambda }^2\\ -2{\lambda }^2& 0\end{array}\right]\left[\begin{array}{cc}0& 2{\lambda }^2\\ -2{\lambda }^2& 0\end{array}\right]=\left[\begin{array}{cc}-4{\lambda }^4& 0\\ 0& -4{\lambda }^4\end{array}\right]=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]$
$A^8=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$